Today I’ll discuss completions in their algebraic context. All this is really a version of Cauchy’s construction of the real numbers, but it’s also useful in algebra, since one can study a ring through its completions (e.g. in algebraic number theory, as I hope to get to soon).

 Generalities on Completions

 Suppose we have a filtered abelian group {G} with a descending filtration of subgroups {\{G_i\}}. Because of this, we can consider “Cauchy sequences” and “convergence:” 

Definition 1

The sequence {\{x_i\} \subset G}, {i \in \mathbb{N}} is Cauchy if for each {A}, there exists {N} large enough that


\displaystyle i,j > N \quad \mathrm{implies} \quad x_i - x_j \in G_A.

The sequence {\{y_i\} \subset G} converges to {y} if for each {A}, there exists {N} large enough that

\displaystyle i>A \quad \mathrm{implies} \quad x_i -y \in G_A.



 If we endow with {G} with the pseudo-metric

\displaystyle d(x,y) := 2^{-\mathrm{ max } \ \{i: x -y \in G_i\}},

then Cauchy sequences and convergence in this language reduce to the corresponding notions in the classical language.

As one constructs the reals from the rationals, one can take a completion: 

Definition 2

The completion {\hat{G}} of {G} is defined as the abelian group consisting of equivalence classes of Cauchy sequences {\{x_i\} \subset G}, where


\displaystyle \{x_i\} \sim \{y_i\} \quad \mathrm{if} \quad x_i-y_i \rightarrow 0.

This is a group under pointwise addition.


 {G} is said to be complete if {G \simeq \hat{G}}; this is equivalent to the assertion that each Cauchy sequence has a unique limit in {G}. Note that the existence of a limit is not enough; the canonical map {G \rightarrow \hat{G}} sending {x \in G} to the equivalence class of the sequence {(x,x, \dots)} is injective if and only if {G} is Hausdorff.

I like formulating the definition in terms of Cauchy sequences, but there is also a more algebraic construction in terms of inverse limits. Given a sequence of groups {H_i, i \in \mathbb{N}}, with homomorphisms {h_{ij}: H_i \rightarrow H_j} for {i>j} such that {h_{ik} = h_{jk} \circ h_{ij} } for {i>j>k}, we define the inverse limit {\varprojlim H_i} to consist of the subset of {\prod H_i} consisting of strings {\{a_i \in H_i\}} such that {h_{ij}(a_i)=a_j}. This is actually a group. If the {H_i} are rings and the {h_{ij}} ring-homomorphisms, {\varprojlim H_i} is a ring too.

The inverse limit construction is perhaps best thought of in more generality, as an example of a categorical limit. Then inverse limits can be defined in arbitrary categories, although they need not exist. 

Proposition 3 We have {\hat{G} \simeq \varprojlim G/G_i}, where the inverse limit is taken to the canonical maps {G/G_i \rightarrow G/G_j} for {i>j}.

 This can be checked directly. Fix {j}. Given a Cauchy sequence {\{x_n\} \subset G}, the images {\overline{x_n} \in G/G_j} eventually stabilize to some {\overline{y_j} \in G/G_j}, so one gets an element {y = \prod \overline{y_j} \in \varprojlim G/G_i}. For the reverse, given an element {y \in \varprojlim G/G_i}, pick liftings {x_n} of {y_n \in G/G_n} to {G} to get a Cauchy sequence. Since we are working modulo equivalence classes, it follows that we have an isomorphism.

Some books seem to define the completion just using the inverse limit and do everything more or less algebraically, but I think it is also useful to think topologically, which can make proofs easier; this tends to be emphasized in material on the {p}-adic numbers.

All the same, it is of course also useful to think algebraically, and the next result is perhaps best thought of using the definition {\hat{G} := \varprojlim G/G_i}. The next result will be useful in proving that completion preserves Noetherianness for rings:

Proposition 4 There is a filtration {\hat{G}_i \subset \hat{G}} with {\hat{G}/\hat{G_i} \simeq G/G_i}. In the resulting topology, {\hat{G}} is complete.

 Define {\hat{G}_i} to consist of elements {\{x_n\}} of the inverse limit with {x_n = 0} for {n \leq i}; this is implied by {x_i = 0 \in G/G_i}. The map {\hat{G}/\hat{G}_i \rightarrow G/G_i} is just projection on the {i}th coordinate. It is easily seen to be both surjective and injective.

Since {\hat{G} \simeq \varprojlim G/G_i \simeq \varprojlim \hat{G}/\hat{G}_i}, we find that {\hat{G} \simeq \hat{\hat{G}}}, so {\hat{G}} is complete.

Note that the result implies moreover {\hat{G}_i/\hat{G}_j \simeq G_i/G_j}.

I didn’t get as far as I would have hoped in this post, but I will keep going next week on completions of rings.